Adding fractions in mixed number form 2 3 25 + 1 3 10 = 2 3 4 + 3 2

Adding fractions in mixed number form
Carefully read the following methods, and make sure you understand the examples:
ο‚· Method 1: Convert all mixed numbers to top-heavy (improper) fractions, then use your usual
method for adding (find a common denominator, then add numerators).
ο‚· Method 2: Write each mixed number separately as whole number plus fraction, then add the
whole numbers separately, add the fractions separately then combine.
Eg:
2
7
3 +4
5
10
Using method 1:
Converting to top-heavy fractions:
2 17
7
47
3 =
π‘Žπ‘›π‘‘ 4
=
5
5
10 10
Finding a common denominator and adding:
17 47 34 47 πŸ–πŸ
+
=
+
=
5 10 10 10 𝟏𝟎
Using method 2:
Splitting whole number and fraction parts:
2
7
2
7
3 +4
=3+ +4+
5
10
5
10
Adding the whole numbers and the fractions separately:
2 7
4
7
11
1
3 + 4 = 7 π‘Žπ‘›π‘‘ +
=
+
=
=1+
5 10 10 10 10
10
Adding the two parts back together at the end:
1
1
𝟏
7+1+
=8+
=πŸ–
10
10
𝟏𝟎
81
1
Note that and 8 are just different ways of writing the same number (It’s also 8.1).
10
10
1. Use method 1 to answer the following questions. You may leave your answer as a top-heavy fraction,
but you should simplify as far as possible.
2
3
3
+1
=
25
10
3
2
2 +3 =
4
5
2. Use method 2 to answer the following questions. You should leave your answer in mixed number
form. The fraction part should be fully simplified.
1
2
6 +7 =
4
7
5
4
20 + 34 =
6
9
Mixed Numbers
Mixed Numbers
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 is not a sensible form, so convert to 6 .
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 is not a sensible form, so convert to 6 .
Challenge:
Challenge:
3
3
1
3
4 βˆ’2 =
3
4
3
3
1
3
4 βˆ’2 =
3
4
Mixed Numbers
Mixed Numbers
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 3 is not a sensible form, so convert to 6 3.
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 3 is not a sensible form, so convert to 6 3.
Challenge:
Challenge:
1
3
4 βˆ’2 =
3
4
1
3
4 βˆ’2 =
3
4
Adding fractions in mixed number form SOLUTIONS
Carefully read the following methods, and make sure you understand the examples:
ο‚· Method 1: Convert all mixed numbers to top-heavy (improper) fractions, then use your usual
method for adding (find a common denominator, then add numerators).
ο‚· Method 2: Write each mixed number separately as whole number plus fraction, then add the
whole numbers separately, add the fractions separately then combine.
Eg:
2
7
3 +4
5
10
Using method 1:
Converting to top-heavy fractions:
2 17
7
47
3 =
π‘Žπ‘›π‘‘ 4
=
5
5
10 10
Finding a common denominator and adding:
17 47 34 47 πŸ–πŸ
+
=
+
=
5 10 10 10 𝟏𝟎
Using method 2:
Splitting whole number and fraction parts:
2
7
2
7
3 +4
=3+ +4+
5
10
5
10
Adding the whole numbers and the fractions separately:
2 7
4
7
11
1
3 + 4 = 7 π‘Žπ‘›π‘‘ +
=
+
=
=1+
5 10 10 10 10
10
Adding the two parts back together at the end:
1
1
𝟏
7+1+
=8+
=πŸ–
10
10
𝟏𝟎
81
1
Note that and 8 are just different ways of writing the same number (It’s also 8.1).
10
10
1. Use method 1 to answer the following questions. You may leave your answer as a top-heavy fraction,
but you should simplify as far as possible.
2
3
3
53 13 212 130 342 πŸπŸ•πŸ
+1
=
+
=
+
=
=
25
10 25 10 100 100 100
πŸ“πŸŽ
3
2 11 17 55 68 πŸπŸπŸ‘
2 +3 =
+
=
+
=
4
5
4
5
20 20
𝟐𝟎
2. Use method 2 to answer the following questions. You should leave your answer in mixed number
form. The fraction part should be fully simplified.
1
2
1 2
7
8
15
πŸπŸ“
6 + 7 = (6 + 7) + ( + ) = (13) + ( + ) = (13) + ( ) = πŸπŸ‘
4
7
4 7
28 28
28
πŸπŸ–
5
4
5 4
15 12
27
20 + 34 = (20 + 34) + ( + ) = 54 + ( + ) = 54 + ( )
6
9
6 9
18 18
18
3
1
𝟏
= 54 + ( ) = 54 + (1 + ) = πŸ“πŸ“
2
2
𝟐
Mixed Numbers SOLUTIONS
Mixed Numbers SOLUTIONS
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 is not a sensible form, so convert to 6 .
3
3
3
6
15
3
𝟐𝟏
1) 5 25 + 6 10 = 11 + 50 + 50 = 𝟏𝟏 πŸ“πŸŽ
3
2
15
8
3
23
3
2
3
10
21
πŸ‘
3
7
3
πŸ“
6
15
3
𝟐𝟏
2
15
8
3
23
πŸ‘
10
3
2
3
10
21
3
7
πŸ“
3) 1 30 + 5 6 = 6 + 6 + 6 = πŸ” πŸ”
41
πŸπŸ‘
4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = πŸ– πŸπŸ–
6
3
2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = πŸ– 𝟐𝟎
3) 1 30 + 5 6 = 6 + 6 + 6 = πŸ” πŸ”
20
3
3
1) 5 25 + 6 10 = 11 + 50 + 50 = 𝟏𝟏 πŸ“πŸŽ
2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = πŸ– 𝟐𝟎
10
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 is not a sensible form, so convert to 6 .
7
13
20
41
πŸπŸ‘
4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = πŸ– πŸπŸ–
𝟏
5) 1 6 + 9 12 = 10 + 12 + 12 = 10 + 12 = 𝟏𝟏 𝟏𝟐
Challenge:
1
3
4 βˆ’2 =
3
4
1
3
1 3
(4 + ) βˆ’ (2 + ) = 2 + ( βˆ’ )
3
4
3 4
4
9
5
πŸ•
=2+( βˆ’ )=2βˆ’
=𝟏
12 12
12
𝟏𝟐
6
7
13
𝟏
5) 1 6 + 9 12 = 10 + 12 + 12 = 10 + 12 = 𝟏𝟏 𝟏𝟐
Challenge:
1
3
4 βˆ’2 =
3
4
1
3
1 3
(4 + ) βˆ’ (2 + ) = 2 + ( βˆ’ )
3
4
3 4
4
9
5
πŸ•
= 2+( βˆ’ )=2βˆ’
=𝟏
12 12
12
𝟏𝟐
Mixed Numbers SOLUTIONS
Mixed Numbers SOLUTIONS
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 3 is not a sensible form, so convert to 6 3.
You can use any method you like to answer these.
If the fraction part of your answer is top-heavy,
you will need to turn it into a mixed number.
4
1
Eg: 5 3 is not a sensible form, so convert to 6 3.
1) 5 25 + 6 10 = 11 + 50 + 50 = 𝟏𝟏 πŸ“πŸŽ
1) 5 25 + 6 10 = 11 + 50 + 50 = 𝟏𝟏 πŸ“πŸŽ
3
3
3
2
6
15
15
8
3
𝟐𝟏
23
πŸ‘
3
2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = πŸ– 𝟐𝟎
10
3
2
3
10
21
3
7
πŸ“
7
2
6
15
15
8
3
𝟐𝟏
23
πŸ‘
10
3
2
3
10
21
3
7
πŸ“
3) 1 30 + 5 6 = 6 + 6 + 6 = πŸ” πŸ”
41
πŸπŸ‘
4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = πŸ– πŸπŸ–
6
3
2) 2 4 + 5 5 = 7 + 20 + 20 = 7 + 20 = πŸ– 𝟐𝟎
3) 1 30 + 5 6 = 6 + 6 + 6 = πŸ” πŸ”
20
3
13
20
41
πŸπŸ‘
4) 3 4 + 4 14 = 7 + 28 + 28 = 7 + 28 = πŸ– πŸπŸ–
𝟏
5) 1 + 9 = 10 + + = 10 + = 𝟏𝟏
6
12
12
12
12
𝟏𝟐
Challenge:
1
3
4 βˆ’2 =
3
4
1
3
1 3
(4 + ) βˆ’ (2 + ) = 2 + ( βˆ’ )
3
4
3 4
4
9
5
πŸ•
=2+( βˆ’ )=2βˆ’
=𝟏
12 12
12
𝟏𝟐
6
7
13
𝟏
5) 1 + 9 = 10 + + = 10 + = 𝟏𝟏
6
12
12
12
12
𝟏𝟐
Challenge:
1
3
4 βˆ’2 =
3
4
1
3
1 3
(4 + ) βˆ’ (2 + ) = 2 + ( βˆ’ )
3
4
3 4
4
9
5
πŸ•
= 2+( βˆ’ )=2βˆ’
=𝟏
12 12
12
𝟏𝟐